/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Find parametric equations for th... [FREE SOLUTION] | 91影视

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Find parametric equations for the curve, and check your work by generating the curve with a graphing utility. The portion of the parabola \(x=y^{2}\) joining \((1,-1)\) and \((1,1),\) oriented down to up.

Short Answer

Expert verified
Parametric equations: \(x = t^2\), \(y = t\), for \(-1 \leq t \leq 1\). Orientation is from down to up.

Step by step solution

01

Identify the Parametric Equations

The given parabolic equation is \(x = y^2\). We want parametric equations where the parameter \(t\) varies. Since we have \(x = y^2\), we choose \(y = t\), then substitution gives \(x = t^2\). This gives us the parametric equations for the curve: \(x = t^2\) and \(y = t\).
02

Determine the Range of Parameter t

The curve is between the points \((1,-1)\) and \((1,1)\). Since \(y = t\), the values of \(t\) range from \(-1\) to \(1\). Thus, \(t\) should vary from \(-1\) to \(1\) to cover the entire segment as required.
03

Verify the Orientation of the Curve

We need to ensure the curve is oriented from \((1,-1)\) to \((1,1)\), which is in the direction from down to up. Since \(t\) is our parameter and increases from \(-1\) to \(1\), \(y\) also increases from \(-1\) to \(1\), confirming the desired orientation.
04

Check with Graphing Utility

Using a graphing utility, plot the parametric equations \(x = t^2\) and \(y = t\) with \(t\) ranging from \(-1\) to \(1\). Ensure the plot of the curve goes from \((1,-1)\) to \((1,1)\) and observe the parabola segment is oriented from bottom to top.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabola
When we think about parabolas, we often picture a U-shaped curve. A parabola is defined mathematically as a set of points that are equidistant from a point called the focus and a line called the directrix. In this specific problem, the parabola is expressed by the equation:
  • \( x = y^2 \)
Here, the variable \( x \) is dependent on the square of \( y \). This signifies a parabolic relationship where the graph opens sideways rather than the more traditional up or down, given this variation of the equation.
This sideways opening is intrinsic because when the variable is squared, it indicates on which axis the parabola opens.
In this task, we're looking specifically at a segment of this parabola, bound between two points: \((1,-1)\) and \((1,1)\). This creates a section of the parabola that goes from one part of the curve to another. Understanding this is essential, as it helps us see how segments of parabolas can be graphed and explored within certain bounds.
Graphing Utility
Graphing utilities are powerful tools that visualize equations so that we can interpret them more easily. A graphing utility can be anything from a simple graphing calculator to sophisticated software like GeoGebra or Desmos. These tools provide insights that are not immediately obvious from raw equations.
  • It allows us to plot parametric equations, such as \(x = t^2\) and \(y = t\), and directly see the curve's shape.
  • In our case, the utility helps you see the segment of the parabola as \(t\) varies, confirming our calculated conclusions.
Using a graphing utility helps verify the portion and orientation of the curve in the question. When plugging in the parametric equations \(x = t^2\) and \(y = t\), with \(t\) ranging from \(-1\) to \(1\), you can see the orientation of the parabola from top to bottom.
This visual tool is especially helpful when learning, as it ensures you understand every step of the graphing process. Furthermore, it allows you to experiment by adjusting parameters and immediately seeing the effects.
Curve Orientation
Orientation is a crucial concept when analyzing curves. It refers to the direction in which a curve proceeds, which can be particularly important in dynamic analyses where the direction and order matter.
In the given problem, the task is to verify that the parabola segment extends from \((1,-1)\) to \((1,1)\) in a downward to upward manner, effectively moving vertically along the \(y\)-axis direction.
  • This was confirmed by setting \(y = t\), which means as \(t\) increased from \(-1\) to \(1\), \(y\) also moves from \(-1\) to \(1\).
  • Because \(x = t^2\) is always positive, each value of \(y\) corresponds to each change in \(x\).
In practical terms, as you plot the curve using the graphing utility, you should see the graph moving upwards as you increase \(t\), confirming that the orientation is indeed from bottom to top. Curve orientation is not just about seeing where a graph starts and ends, but understanding the order of movement through the parameter.鈥潁]}

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Most popular questions from this chapter

Show that the graph of the given equation is a parabola. Find its vertex, focus, and directrix. $$x^{2}-2 \sqrt{3} x y+3 y^{2}-8 \sqrt{3} x-8 y=0$$

Find the eccentricity and the distance from the pole to the directrix, and sketch the graph in polar coordinates. $$ \text { (a) } r=\frac{4}{2+3 \cos \theta} \quad \text { (b) } r=\frac{5}{3+3 \sin \theta} $$

If \(f^{\prime}(t)\) and \(g^{\prime}(t)\) are continuous functions, and if no segment of the curve $$ x=f(t), \quad y=g(t) \quad(a \leq t \leq b) $$ is traced more than once, then it can be shown that the area of the surface generated by revolving this curve about the \(x\) -axis is $$ S=\int_{a}^{b} 2 \pi y \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ and the area of the surface generated by revolving the curve about the \(y\) -axis is $$ S=\int_{a}^{b} 2 \pi x \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ [The derivations are similar to those used to obtain Formulas (4) and (5) in Section 6.5. ] Use the formulas above in these exercises. The equations \(x=a \phi-a \sin \phi, \quad y=a-a \cos \phi \quad(0 \leq \phi \leq 2 \pi)\) represent one arch of a cycloid. Show that the surface area generated by revolving this curve about the \(x\) -axis is given by \(S=64 \pi a^{2} / 3\).

If \(f^{\prime}(t)\) and \(g^{\prime}(t)\) are continuous functions, and if no segment of the curve $$ x=f(t), \quad y=g(t) \quad(a \leq t \leq b) $$ is traced more than once, then it can be shown that the area of the surface generated by revolving this curve about the \(x\) -axis is $$ S=\int_{a}^{b} 2 \pi y \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ and the area of the surface generated by revolving the curve about the \(y\) -axis is $$ S=\int_{a}^{b} 2 \pi x \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}} d t $$ [The derivations are similar to those used to obtain Formulas (4) and (5) in Section 6.5. ] Use the formulas above in these exercises. By revolving the semicircle $$ x=r \cos t, \quad y=r \sin t \quad(0 \leq t \leq \pi) $$ about the \(x\) -axis, show that the surface area of a sphere of radius \(r\) is \(4 \pi r^{2}\).

Prove: The line tangent to the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ at the point \(\left(x_{0}, y_{0}\right)\) has the equation $$\frac{x x_{0}}{a^{2}}+\frac{y y_{0}}{b^{2}}=1$$

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